Schmitt Trigger: NAND Gates

Discussion in 'Homework Help' started by Steve1992, Oct 9, 2010.

  1. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
    100
    0
    When P is high, its voltage is 4.5 V.
    M must go below 2.5 V to make P go low at 0.5 V.

    Use calculations to show that L has to be taken below 1.1 V to make P go low.


    My calculations:

    4.5V (P) – 1.1V (L) = 3.4V

    3.4V/(330k + 470k) = 4.25uA

    4.25uA*330k = 1.4V

    1.1V (L) + 1.4V = 2.5V = M

    Right?
     
  2. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    You cannot use 1.1V in your calculation because that is the voltage you are supposed to calculate.

    Because of the extremely high input impedance of the CMOS gate, the input current is basically 0. Therefore your problem can be simplified to the finding of an input voltage that gives point 'M' 2.5V.

    Try again.

    [​IMG]
     
  3. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
    100
    0
    (4.5V - 2.5V)/470k = 4.25uA

    2.5V/330k = 7.57uA

    7.57uA - 4.25uA = 3.32uA

    3.32uA*330k = 1.1V:confused:
     
  4. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    Correct.

    This is the only current in the loop. There is no other current path.

    Wrong. The current in 330K is the same as 470K because they are in series. The current in 470K has already shown to be 4.25uA, so current in 330K is also 4.25uA.

    Voltage drop across 330K is thus 4.25uA * 330K = 1.403V

    With M = 2.5V, it follows that the input voltage must be 2.5 - 1.403 = 1.098V.
     
  5. Steve1992

    Thread Starter Senior Member

    Apr 7, 2006
    100
    0
    Thank you.
     
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